Problem: Simplify and expand the following expression: $ \dfrac{r}{5r + 5}-\dfrac{4r}{4r + 7} $
Explanation: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(5r + 5)(4r + 7)$ Multiply the first term by $\dfrac{4r + 7}{4r + 7}$ $ \begin{align*} \dfrac{r}{5r + 5} \times \dfrac{4r + 7}{4r + 7} & = \dfrac{(r)(4r + 7)}{(5r + 5)(4r + 7)} \\ & = \dfrac{4r^2 + 7r}{(5r + 5)(4r + 7)}\end{align*} $ Multiply the second term by $\dfrac{5r + 5}{5r + 5}$ $ \begin{align*} \dfrac{4r}{4r + 7} \times \dfrac{5r + 5}{5r + 5} & = \dfrac{(4r)(5r + 5)}{(4r + 7)(5r + 5)} \\ & = \dfrac{20r^2 + 20r}{(4r + 7)(5r + 5)}\end{align*} $ Now we have: $ = \dfrac{4r^2 + 7r}{(5r + 5)(4r + 7)} - \dfrac{20r^2 + 20r}{(4r + 7)(5r + 5)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{4r^2 + 7r - (20r^2 + 20r)}{(5r + 5)(4r + 7)} $ $ = \dfrac{4r^2 + 7r - 20r^2 - 20r}{(5r + 5)(4r + 7)} $ $ = \dfrac{-16r^2 - 13r}{(5r + 5)(4r + 7)}$ Expand the denominator: $ = \dfrac{-16r^2 - 13r}{20r^2 + 55r + 35}$